0

Guest

Courses New

find the no. of solution of -x^2+x-1 = x^2008 + |x|

find the no. of solution of
-x^2+x-1 = x^2008 + |x|

Grade:12

2 Answers

Pratham Ashish
17 Points
14 years ago

hi rahul,

dis ques can be done by d approach of graph,,,

for LHS,

f = -x^2+x-1       ,f < 0 for all x <0,n f<0 for x>1, f >0 for 0<x>1........

 

since dy/dx < 0, n dis is a monotonus fxn so it decreases from +infinite to o at x=1,,,,,,,,,,

for rhs, f = x^2008 + |x|,   dis is a increasing fxn,,,,,n for all x f>0....

at x=1 , f=2,,at x =0,f=0,,,

so d two fxn can only intersect at one point in b/w 0<x>1...

so no. of soln = 1.

Rajesh Juluru
13 Points
13 years ago

for x>0,

-x^2-1=x^2008

x^2008+x^2+1=0

So no positive roots (since all are positive terms)

for x<0,

-x^2+2x-1=x^2008

x^2008+(x-1)^2=0

so no negative roots (since all are positive terms)

So i think number of solutions is zero

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More